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求f(x,y)=sinxy/√(x2+y2) ,(x2+y2≠0),f(0,0)=0,在R2上连续性...

设x=rcosu,y=rsinu,则limf(x,y)=limsin[r^2*cosusinu]/r→0,∴f(x,y)在(0,0)处连续,于是f(x,y)在R2上连续.

首先单变元用洛必达则易知lim xlnx=0x趋于0 于lim (x^2+y^2)*ln(x^2+y^2)=0(xy)趋于(00) |xy|*ln(x^2+y^2)感觉提问主意不是很清晰 这里的只能参考了

对?ε>0,取δ=2ε,当(x,y)属于(0,0)的δ邻域U(δ),即 x2+y2 |f(x,y)?f(0,0)|=|x2y| x2+y2 ≤|x| 2 ≤ x2+y2 2 于是,f(x,y)在原点处连续;由f′x(0,0)=lim △x→0 f(0+△x,0) △x =0及f′y(0,0)=lim △y→0 f(0,0+△y) △y =0,知f(x,y)在原点处的两个偏导数存

令x=ρcosθy=ρsinθ,则limx,y→0f(x,y)=limρ→0+12ρ2sin2θlnρ2=sin2θlimρ→0+lnρρ-2=sin2θlimρ→0+1-2ρ-4=0=f(0,0)∴f(x,y)在原点(0,0)处连续limx→0f(x,0)-f(0,0)x=limx→00x=0=fx(0,0),同理fy(0,0)=0即f(x,y)在(0,0)点处的两个一阶偏导数都存在且limρ→0△z-fx′(0,0)△x-fx′(0,0)△y(△x)2+(△y)2=limρ→012ρ2sin2θlnρ2ρ=limρ→0sin2θρlnρ=0∴f(x,y)在原点(0,0)处可微.

对ε>0,取δ=2ε,当(x,y)属于(0,0)的δ邻域U(δ),即x2+y2

(1)当x2+y2≠0时,f(x,y)在点(x,y)处是连续的;当x2+y2=0时,即(x,y)=(0,0),由于:|f(x,y)f(0,0)|=|xy2x2+y2|≤|y|,所以f(x,y)在点(0,0)处也是连续的.(2)通过偏导

∵lim(x,y)→(0,0)(x2+y)=0,|sin1x2+y2|≤1∴由无穷小与有界函数的乘积依然是无穷小的性质,知lim(x,y)→(0,0)f(x,y)=0=f(0,0)∴f(x,y)在点(0,0)连续.又f′x(0,0)=limx→0f(x,0)f(0,0

∵(x,y)→(0,0)时,|sin 1 x2+y2 |≤1,α>0时, lim (x,y)→(0,0) (x2+y2)α=0∴由“无穷小与有界函数的乘积依然是无穷小”的性质,知当α>0时, lim ,(x,y)→(0,0) f(x,y)=0=f(0,0)又fx(0,0)= lim x→0 f(x,0)?f(0,0) x = lim x→0 x2αsin 1 x2 x = lim x→0 x2α?1? 1 x2 = lim x→0 x2α?3∴当2α-3>0时,即α> 3 2 时,fx(0,0)存在.

:∵f(x)=x+sinx(x∈R),∴f(-x)=-x-sinx=-(x+sinx)=-f(x),即f(x)=x+sinx(x∈R)是奇函数,∵f(y2-2y+3)+f(x2-4x+1)≤0,∴f(y2-2y+3)≤-f(x2-4x+1)=f[-(x2-4x+1)],由f'(x)=1-cosx≥0,∴函数单调递增.∴(y2-2y+3)≤-(x2-4x+

全微分不存在理由如下:f(x,y)对x的偏导,f'x(x,0)=lim[f(x,0)-f(0,0)]/x (x趋向0)可得f'x(x,0)=|x|/x,显然,f(x,y)对x的偏导不存在,因此,全微分不存在

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